I'm trying to show that the Weiestrass's $\sigma(z)$ function is defined by $$\sigma(z)=z\prod_{\omega\in L'}\{(1-\frac{z}{\omega})e^{\frac{z}{\omega}+\frac{z^2}{2\omega^2}}\}.$$
I tried to integrate the Weiestrass's $\zeta(z)$ which is defined by $$\zeta(z)=\cfrac{1}{z}+\sum_{\omega\in L'}\Big(\cfrac{1}{z-\omega}+\cfrac{1}{\omega}+\cfrac{z}{\omega^2}\Big),$$ term-by-term in the way $$\int_{0}^{z}\zeta(u)du$$ but I don't pretty sure if the limits of integration are correct because the corresponding integral to the term 1/z is $$\int_{0}^{z}\cfrac{1}{u}du=\log(u)\Big|_{0}^{z}$$ which is not defined at u=0. Someone could tell me if the limits are correct or if I am missing something. Any help would really grateful.
I assume you've defined $\sigma(z)$ as the unique function whose logarithmic derivative is $\zeta(z)$ and whose first Taylor term is $z$, and then you're trying to prove the product formula for $\sigma(z)$.
You can indefinitely integrate the series for $\zeta(s)$, but then you need to add a different constant of integration to each summand to make the series converge, and the right ones to make it match $\sigma(z)$. (If you want, you can think of this as definitely integrating from $z=1$, since $\ln1=0$.)
It'd be easier, I think, to simply take the logarithmic derivative of the product formula (which has first Taylor term $z$ after expanding) and verify you get $\zeta(z)$. Indeed, if you do this and study how it looks "backwards" you'll know what constant of integration to add to each term when integrating $\zeta(z)$.